2024黑龙江CCPC

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2024黑龙江省赛

B. String

思路:栈模拟
代码：

/*
* @Author: 	Hfuubigstrength
* @email:	2854614012@qq.com
* @Date:   2024-05-28 23:32:45
*/
#include <bits/stdc++.h>
//#define int long long
#define PII pair<int,int>
#define LL long long
#define fi first
#define se second
#define debug(a) cout<<#a<<"="<<a<<endl;
#define all(x) (x).begin(),(x).end()
#define pb push_back
#define sz(x) (int)x.size()
using namespace std;string s;
signed main(){ios::sync_with_stdio(false);cin.tie(0);cin >> s;stack<char>stk;for(auto it : s){stk.push(it);if(stk.size() >= 3){char a = stk.top(); stk.pop();char b = stk.top(); stk.pop();char c = stk.top(); stk.pop();if(!(a == b && b == c)){stk.push(c);stk.push(b);stk.push(a);}}}string ans;while(stk.size()){ans += stk.top();stk.pop();}reverse(all(ans));if(!ans.size()) cout << "NAN";else cout << ans;return 0;
}

D. Card Game

思路：贪心，把A从大到小排序，把B不为-1的从小到大排序，按顺序出牌，这种方法最优
代码：

/*
* @Author: 	Hfuubigstrength
* @email:	2854614012@qq.com
* @Date:   2024-05-28 23:42:10
*/
#include <bits/stdc++.h>
// #define int long long
#define PII pair<int,int>
#define LL long long
#define fi first
#define se second
#define debug(a) cout<<#a<<"="<<a<<endl;
#define all(x) (x).begin(),(x).end()
#define pb push_back
#define sz(x) (int)x.size()
using namespace std;const int N = 100010;
int a[N], b[N], ah, bh, n;void solve(){cin >> n >> ah >> bh;for(int i = 1; i <= n; i ++ ) cin >> a[i];vector<int>bb;int cntb = 0;for(int i = 1; i <= n; i ++ ){cin >> b[i];if(b[i] != -1) bb.pb(b[i]);else cntb ++;} sort(all(bb));while(cntb -- ){bb.pb(-1);}sort(a + 1, a + 1 + n, greater<int>());for(int i = 1; i <= n; i ++ ){if(bb[i - 1] >= 0 && a[i] != -1) ah -= bb[i - 1];if(a[i] >= 0 && bb[i - 1] != -1) bh -= a[i];if(ah > 0 && bh <= 0){cout << "yes\n";return;}if(ah <= 0){cout << "no\n";return;}}cout << "no\n";
}signed main(){ios::sync_with_stdio(false);cin.tie(0);int tt = 1;cin >> tt;while(tt -- ){solve();}return 0;
}

F. Photography

思路：先预处理每一个点最大的三个邻接点，存下标。然后暴力枚举每两条边，如果能形成一条链，就从链的两端进行拓展出最大的那个点，这是处理长度大于3的情况。还要注意考虑长度为1, 2的情况。这题有孤点。

/*
* @Author: 	Hfuubigstrength
* @email:	2854614012@qq.com
* @Date:   2024-05-29 03:30:40
*/
#include <bits/stdc++.h>
//#define int long long
#define PII pair<int,int>
#define LL long long
#define fi first
#define se second
#define debug(a) cout<<#a<<"="<<a<<endl;
#define all(x) (x).begin(),(x).end()
#define pb push_back
#define sz(x) (int)x.size()
using namespace std;const int N = 10010;
int a[N], big[N][3];
int n, m, ans, now;
vector<PII>q[N];
vector<PII>edges;	int work(int same1, int same2, int diff1, int diff2){int four = -1;int now = a[same1] + a[diff1] + a[diff2];bool f = 1;if(big[diff1][0] == same1 || big[diff1][0] == diff2 || big[diff1][0] == -1){if(big[diff1][1] == same1 || big[diff1][1] == diff2 || big[diff1][1] == -1){if(big[diff1][2] == -1 || big[diff1][2] == diff2 || big[diff1][2] == same1){f = 0;}else now += a[big[diff1][2]], four = big[diff1][2];}else now += a[big[diff1][1]], four = big[diff1][1];}else now += a[big[diff1][0]], four = big[diff1][0];if(big[diff2][0] == same1 || big[diff2][0] == diff1 || big[diff2][0] == four || big[diff2][0] == -1){if(big[diff2][1] == same1 || big[diff2][1] == diff1 || big[diff2][1] == four || big[diff2][1] == -1){if(big[diff2][2] == same1 || big[diff2][2] == diff1 || big[diff2][2] == four || big[diff2][2] == -1){f = 0;}else now += a[big[diff2][2]];}else now += a[big[diff2][1]];}else now += a[big[diff2][0]];return now;
}signed main(){ios::sync_with_stdio(false);cin.tie(0);cin >> n >> m;for(int i = 1; i <= n; i ++ ){cin >> a[i];for(int j = 0; j <= 2; j ++ ) big[i][j] = -1;}for(int i = 1; i <= m; i ++ ){int x, y; cin >> x >> y;q[x].pb({a[y], y});q[y].pb({a[x], x});edges.pb({x, y});if(x != y) ans = max(ans, a[x] + a[y]);}for(int i = 1; i <= n; i ++ ) ans = max(ans, a[i]);for(int i = 1; i <= n; i ++ ){sort(all(q[i]));int len = q[i].size();if(len >= 1) big[i][0] = q[i][len - 1].se;if(len >= 2) big[i][1] = q[i][len - 2].se;if(len >= 3) big[i][2] = q[i][len - 3].se;}for(int i = 0; i < m; i ++ ){for(int j = 0; j < m; j ++ ){if(i == j) continue;int x = edges[i].fi, y = edges[i].se, xx = edges[j].fi, yy = edges[j].se;if(x == xx) ans = max(ans, work(x, xx, y, yy));if(x == y) ans = max(ans, work(x, y, xx, yy));if(y == xx) ans = max(ans, work(y, xx, x, yy));if(y == yy) ans = max(ans, work(y, yy, x, xx));}}cout << ans << endl;return 0;
}

I. This is an easy problem

思路：模拟
代码

/*
* @Author: 	Hfuubigstrength
* @email:	2854614012@qq.com
* @Date:   2024-05-28 23:26:33
*/
#include <bits/stdc++.h>
//#define int long long
#define PII pair<int,int>
#define LL long long
#define fi first
#define se second
#define debug(a) cout<<#a<<"="<<a<<endl;
#define all(x) (x).begin(),(x).end()
#define pb push_back
#define sz(x) (int)x.size()
using namespace std;int n;signed main(){ios::sync_with_stdio(false);cin.tie(0);cin >> n;int cnt = 0;while(n){cnt += n % 2;n /= 2;}cout << cnt;return 0;
}

J. Trade

思路：dp, $d p [i] [j]$ 表示，走到第 $i, j$ 的路径上的 $\sum{b[i][j]}$ ，转移方式类似摘花生那题并要结合条件判断当前点是否是利润非负。最后看最后一行或者最后一列有没有小于无穷大的数即可。
代码

/*
* @Author: 	Hfuubigstrength
* @email:	2854614012@qq.com
* @Date:   2024-05-29 00:46:32
*/
#include <bits/stdc++.h>
//#define int long long
#define PII pair<int,int>
#define LL long long
#define fi first
#define se second
#define debug(a) cout<<#a<<"="<<a<<endl;
#define all(x) (x).begin(),(x).end()
#define pb push_back
#define sz(x) (int)x.size()
using namespace std;const int N = 1010;
LL a[N][N], b[N][N], dp[N][N];
int n, m;signed main(){ios::sync_with_stdio(false);cin.tie(0);cin >> n >> m;for(int i = 1; i <= n; i ++ ){for(int j = 1; j <= m; j ++ ){cin >> a[i][j];dp[i][j] = 9e18;}}for(int i = 1; i <= n; i ++ ){for(int j = 1; j <= m; j ++ ){cin >> b[i][j];}}dp[1][1] = b[1][1];for(int i = 1; i <= n; i ++ ){for(int j = 1; j <= n; j ++ ){if(i == 1 && j == 1) continue;if(i >= 2) dp[i][j] = min(dp[i - 1][j], dp[i][j]);if(j >= 2) dp[i][j] = min(dp[i][j - 1], dp[i][j]);dp[i][j] += b[i][j];if(a[i][j] - a[1][1] - dp[i][j] < 0) dp[i][j] = 9e18;}}bool f = 0;for(int i = 1; i <= n; i ++ ){if(dp[i][m] < 9e18) f = 1;}for(int i = 1; i <= m; i ++ ){if(dp[n][i] < 9e18) f = 1;}if(f) cout << "YES";else cout << "NO";return 0;
}

K. Puzzle

思路：纯模拟
代码：

#include <bits/stdc++.h>
//#define int long long
#define PII pair<int,int>
#define LL long long
#define fi first
#define se second
#define debug(a) cout<<#a<<"="<<a<<endl;
#define all(x) (x).begin(),(x).end()
#define pb push_back
#define sz(x) (int)x.size()
using namespace std;int a[10];
set<int>s;
signed main(){ios::sync_with_stdio(false);cin.tie(0);cin >> a[1] >> a[2] >> a[3] >> a[4];sort(a + 1, a + 5);set<int>s;do{for(int i = 1; i <= 3; i ++ ){for(int j = 1; j <= 3; j ++ ){for(int k = 1; k <= 3; k ++ ){if(i == 1 && j == 1 && k == 1) s.insert(a[1] + a[2] +a[3] + a[4]);if(i == 1 && j == 1 && k == 2) s.insert(a[1] + a[2] +a[3] - a[4]);if(i == 1 && j == 1 && k == 3) s.insert(a[1] + a[2] +a[3] * a[4]);if(i == 1 && j == 2 && k == 1) s.insert(a[1] + a[2] -a[3] + a[4]);if(i == 1 && j == 2 && k == 2) s.insert(a[1] + a[2] -a[3] - a[4]);if(i == 1 && j == 2 && k == 3) s.insert(a[1] + a[2] -a[3] * a[4]);if(i == 1 && j == 3 && k == 1) s.insert(a[1] + a[2] *a[3] + a[4]);if(i == 1 && j == 3 && k == 2) s.insert(a[1] + a[2] *a[3] - a[4]);if(i == 1 && j == 3 && k == 3) s.insert(a[1] + a[2] *a[3] * a[4]);if(i == 2 && j == 1 && k == 1) s.insert(a[1] - a[2] +a[3] + a[4]);if(i == 2 && j == 1 && k == 2) s.insert(a[1] - a[2] +a[3] - a[4]);if(i == 2 && j == 1 && k == 3) s.insert(a[1] - a[2] +a[3] * a[4]);if(i == 2 && j == 2 && k == 1) s.insert(a[1] - a[2] -a[3] + a[4]);if(i == 2 && j == 2 && k == 2) s.insert(a[1] - a[2] -a[3] - a[4]);if(i == 2 && j == 2 && k == 3) s.insert(a[1] - a[2] -a[3] * a[4]);if(i == 2 && j == 3 && k == 1) s.insert(a[1] - a[2] *a[3] + a[4]);if(i == 2 && j == 3 && k == 2) s.insert(a[1] - a[2] *a[3] - a[4]);if(i == 2 && j == 3 && k == 3) s.insert(a[1] - a[2] *a[3] * a[4]);if(i == 3 && j == 1 && k == 1) s.insert(a[1] * a[2] +a[3] + a[4]);if(i == 3 && j == 1 && k == 2) s.insert(a[1] * a[2] +a[3] - a[4]);if(i == 3 && j == 1 && k == 3) s.insert(a[1] * a[2] +a[3] * a[4]);if(i == 3 && j == 2 && k == 1) s.insert(a[1] * a[2] -a[3] + a[4]);if(i == 3 && j == 2 && k == 2) s.insert(a[1] * a[2] -a[3] - a[4]);if(i == 3 && j == 2 && k == 3) s.insert(a[1] * a[2] -a[3] * a[4]);if(i == 3 && j == 3 && k == 1) s.insert(a[1] * a[2] *a[3] + a[4]);if(i == 3 && j == 3 && k == 2) s.insert(a[1] * a[2] *a[3] - a[4]);if(i == 3 && j == 3 && k == 3) s.insert(a[1] * a[2] *a[3] * a[4]);}}}}while(next_permutation(a + 1,a + 1 + 4));cout << sz(s);return 0;
}